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On 5/21/24 1:52 AM, olcott wrote:On 5/21/2024 3:05 AM, Mikko wrote:On 5/20/2024 10:37 PM, Richard Damon wrote:Not ALLOWED. p is DEFINED to be something, so it is that/.On 5/20/24 10:56 PM, olcott wrote:>On 5/20/2024 9:24 PM, Richard Damon wrote:>On 5/20/24 9:54 PM, olcott wrote:>On 5/20/2024 7:57 PM, Richard Damon wrote:>On 5/20/24 2:59 PM, olcott wrote:>On 5/19/2024 6:30 PM, Richard Damon wrote:>On 5/19/24 4:12 PM, olcott wrote:>On 5/19/2024 12:17 PM, Richard Damon wrote:>On 5/19/24 9:41 AM, olcott wrote:>>>
True(L,x) is always a truth bearer.
when x is defined as True(L,x) then x is not a truth bearer.
So, x being DEFINED to be a certain sentence doesn't make x to have the same meaning as the sentence itself?
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What does it mean to define a name to a given sentence, if not that such a name referes to exactly that sentence?
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p = ~True(L,p) // p is not a truth bearer because its refers to itself
Then ~True(L,p) can't be a truth beared as they are the SAME STATEMENT, just using different "names".
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Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
p = ~True(L,p) Truthbearer(L,p) is false
q = ~True(L,p) Truthbearer(L,q) is true
Irrelvent.
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If Truthbearer(L, p) is FALSE, and since p is just a NAME for the statement ~True(L, p), that means that True(L. p) is not a truth bearer and True has failed to be the required truth predicate.
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That is the same thing as saying that
True(English, "this sentence is not true") is false
proves that True(L,x) is not a truthbearer.
Nope, why do you say that?
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What logic are you even TRYING to use to get there?
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I think you don't understand what defining a label to represent a statement means.
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I did not said the above part exactly precisely to address
your objection.
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p is defined as ~True(L,p)
LP is defined as "this sentence is not true" in English.
Thus True(L,p) ≡ True(English,LP) and
Thus True(L,~p) ≡ True(English,~LP)
So, you admit that you did not answer the problem.
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And that you think Strawmen and Red Herring are valid forms of logic.
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How does p defined as ~True(L, p) NOT generate the shown contradiction when you begin by saying True(L, p) must not be true (and thus false) because p has not chain to truthbears?
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p := ~True(L, p) is false
p := ~True(L, ~p) is false
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p is tossed out on its ass as a type mismatch error for every system
of bivalent logic before it gets any chance to be evaluated in any
other way.
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